Question 25

[yfangl097543]

I'm not sure if the calculations in the explanation is correct, since the question says that 3/4 of the height is submerged, not 3/4 of the total mass of the ball.

[admin]

The calculation is correct.

[yfangl097543]

This doesn't seem to make sense to me, because of Archimede's principle:

The buoyant force is equal to the weight of fluid displaced, so since the ball is not moving up or down, the weight of the ball must be equal to the buoyant force.

Let's say that the volume of the ball is 5 units, so since 3/4 of the HEIGHT of the ball is submerged, the volume submerged must be more than 3/4 of 5 because of the shape of the ball. So let's say that 4 units of the ball's volume is submerged as an approximation.

Therefore, F(buoyant) = Weight of ball, so

volume of fluid X displaced (same of volume of ball submerged) * density of fluid X * gravity = 5 units * density of ball (determined to be .5) * gravity, so

4 * d of fluid X = 5 * .5,

d of fluid X = 2.5 / 4 = .625

Therefore, unless 3/4 of the VOLUME of the ball is submerged, the density of the fluid cannot be .67

Am I missing something?

[admin]

No, you're not missing something, the physics you used is good.

It is true that normally on the MCAT the object immersed is a block. That way, the rule of thumb regarding SG being the fraction of the height submerged is precise. However, given your approximations, you have shown that it is also a reasonable approximation even if the shape is a sphere (the answer being 0.63 g/cc vs 0.67 g/cc with no answer choices even close).

For those who wish to see the proof regarding a block or cube:

Let h = the height of the block
Let x = the height of the block above the surface of the fluid
Let A = the area of the bottom surface of the block

The volume of displaced water is the volume of the object beneath the surface of the water:

V = (h - x) A

As you mentioned, according to Archimedes' principle, the upward buoyant force on the block is the weight of the water displaced:

Fb = W(water) = D(water)g(h - x)A

where D(water) is the density of water. This must balance the weight W of the block:

Fb = W = DgAh

where D is the density of the block. Thus:

D(water)g(h - x)A = DgAh

Cancel "g" and "A" on both sides and rearrange the equation. Now we have the fractional height of the block above the surface of water given by:

x/h = 1 - SG

where SG = D/D(water)


This link below does not have the proof as above but it has an example of its use:

http://webspinners.com/dlblanc/tectonic/floating.php

[dnpgr16513]

Why do you flip 3/4 to 4/3? Please help me!

THANKS

[jellywing_2058]

This is basic math. See the following:

ρball/ρl = 3/4
ρl/pball = 4/3
pl = pball (4/3)
pl = 0.5 g/cm3 x 4/3
pl = 1/2 g/cm3 x 4/3
pl = 2/3 g/cm3
pl = 0.67 g/cm3.

[dnpgr16513]

Ok, great! That's not too bad..

[hua8986059]

I've noticed that the stuff that MCAT likes to stress are the stuff that actually are very medically applicable. I worked at a cardiologist convention where they talked about ideal fluid flow etc... good stuff to learn